1. A 1.0kg mass is lifted vertically two meters (on Earth) with a constant 2.3m/s 2 acceleration. How much work is done by the lifting force, F?

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Emmeline Mathews
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1 Five Easy Pieces 1. A 1.0kg mass is lifted vertically two meters (on Earth) with a constant 2.3m/s 2 acceleration. How much work is done by the lifting force, F? (a) W = 19.6J (b) W = 19.6J (c) W = 24.2J (d) W = 24.2J 2. An 80kg man has an apparent weight of 1504N at the bottom of a circular dip. If his speed is 6.0m/s, what is the radius of the circle? (a) r = 9m (b) r = 1.26m (c) r = 4m (d) r = 9.8m 3. A 700kg car is traveling with a speed of 30m/s. If 5.2s later, its speed is 19.6m/s, how much work, in total, was done to the car? Note the use of kj = kilojoules to make the numbers smaller. (a) W = 1770kJ (b) W = 35kJ (c) W = 9.8kJ (d) W = 180kJ

2 4. A 5.0kg mass is being held against a vertical wall as shown. If the coefficient of static friction between the mass and the wall is.64, what is the minimum horizontal force P needed to hold mass in place? M P (a) P = 49N (b) P = 76.6N (c) P =.64N (d) P = 31.36N 5. A 5kg mass and a 3kg mass are connected by a massless rope on a horizontal, frictionless surface. The masses are made to accelerate at 1m/s 2 to the right as shown. In which case will the rope s tension have a larger magnitude? 1m/s 2 1m/s 2 3kg 5kg 5kg 3kg ase #1 ase #2 (a) ase # 1 (b) ase # 2 (c) The tension is the same in both cases. (d) It is impossible to determine.

3 Work, Work, Work Questions 6 through 10 refer to the following setup. Émilie duhâtelet dropsa.25kg ball fromrest, 1.6mabove asandpit. 6. If the ball hits the sand going 3.2m/s, how much work was done by air resistance during the ball s fall? (a) 1.28J (b) 3.92J (c) 2.64J (d) 1.28J 7. Ignoring gravity, how much work must the sand do in order to stop the 3.2 m/s ball? (a) 1.28J (b) 3.92J (c) 2.64J (d) 1.28J

4 8. If the force exerted by the sand increases linearly with depth below the sand, as shown on the graph below (y = 0 is at the top of the sand pit and down is positive), how far will the ball sink below the surface before stopping? Again, ignore gravity in this calculation. F(N) 35.1 y(m) (a).037m (b).191m (c).086m (d).060m 9. If the sand provides 24Watt of average power, estimate the time it takes for the ball to stop. (a).053s (b).025s (c) 9.8s (d) 3s 10. If we were to include the effect of gravity in the previous calculation, the ball would: (a) go deeper into the sand. (b) go the same distance into the sand. (c) go less distance into the sand. (d) bounce off the sand.

5 Superman Returns (to lift things) Questions 11 through 15 refer to the following setup and figure. The two gigantic globes (mass M 1 = 765kg and M 2 = 500kg respectively) from the top of the Daily Planet Headquarters have fallen down to the ground, and Superman has arrived to clean up for us (again). Miraculously, both globes are still round (in shape), lying as shown below, and have the same radius. Superman stretches to warmup and then applies a 3000N horizontal force to M 2. In all questions ignore friction. M 1 M N 40

6 11. The instant Superman applies his horizontal force, which of the following is the correct free body diagram for M 2? (Unlike Mastering Physics, I only care about direction!) In each case, n g2 is the normal force due to the ground while is the normal force due to the contact between M 1 and M 2. (a) (b) 3000N 3000N W 2 W 2 n g2 n g2 (c) (d) 3000N 3000N W 2 W 1 W 2 n g2 n g2 12. If Superman s force is unable to make the globes move, what is the magnitude of the force? (a) = 6000N (b) = 7900N (c) = 4900N (d) = 3000N 13. How large is the normal force due to the ground, n g2? (a) 12397N (b) 7900N (c) 4900N (d) 3000N

7 14. Which of thefollowing is thecorrect free bodydiagramform 1? In each case, n g1 is the normal force due to the ground, n I is the normal force due to the incline, is the normal force due to the contact between M 1 and M 2. (a) 3000N (b) W 1 n I W 1 n g1 n g1 (c) (d) n I W 1 n I W 1 n g1 n g1 15. From M 1 s free body diagram, we can conclude that: (a) n g1 = W 1 (b) n g1 > W 1 (c) n g1 < W 1 (d) n g1 and W 1 are unrelated.

8 Still Lifting Questions 16 through 20 refer to the following setup and figure. Tiredofdealingwiththetwoglobes, SupermanuseshisX-rayvision(or whatever it s called) to melt them together to form one M = 1200kg, rectangular mass. Being Super-persistent, Superman continues to apply a horizontal force, P, to the mass even after it starts to slide up the incline. v M P 40

9 16. If we include friction, which of the following is the correct free body diagram for M? (Unlike Mastering Physics, I always split weight into components on an incline.) (a) fk n (b) fk W P P W W n W (c) fk n (d) fk n P P W W W W 17. Which of the following graphs shows how to correctly split P into its perpendicular and parallel components? (a) (b) P cos40 40 P sin40 40 P sin40 P cos40 (c) P sin40 (d) P cos P cos40 P sin50

10 18. Which of the following is the correct listing of the parallel and perpendicular weight of the M = 1200kg mass on the 40 incline? Note the use of kn = kilonewtons to make the numbers smaller. (a) W = 9.0kN, W = 7.6kN (b) W = 7.6kN, W = 9.0kN (c) W = 11.8kN, W = 9.0kN (d) W = 0kN, W = 11.8kN 19. If P = 21kN, what is the magnitude of the normal force acting on the mass? (a) n = 9.0kN (b) n = 7.6kN (c) n = 22.5kN (d) n = 21kN 20. If P = 21kN and µ k =.35, what is the mass s acceleration? Use up the incline as the positive direction. (a) a =.5m/s 2 (b) a = 1.7m/s 2 (c) a = 17.5m/s 2 (d) a =.26m/s 2

11 21. The Great Glass Elevator One day finds little harlie Bucket (mass 48kg) and Willy Wonka riding around (on Earth) in their fabulous great glass elevator. If you ve never read the book, the great glass elevator is an elevator that can move in any direction you might wish. Sometime during their trip, little harlie Bucket and Willy Wonka turn a corner in the great glass elevator by zooming around a 100m radius circle with a speed of 22.2m/s. For reasons that only make sense to Willy Wonka (and your instructor), harlie is riding in the elevator standing on a scale. enter of ircle 100m (a) If the coefficient of static friction between harlie Bucket and his scale is.39, will he be able to remain not-sliding as he travels around this circle? Assume, as shown above, that the center of the circle is directly to the right of harlie Bucket. (You must do a calculation of some sort to get full credit on this problem which is why there is an extra page provided.) (10pts)

13 (b) By fiddling with some buttons, Willy Wonka discovers that he can tilt the great glass elevator to any angle that he wishes. To what angle θ should Willy Wonka tilt the elevator so that no friction is necessary for harlie Bucket to go around a 100m radius circle (whose center is still directly to his right) with a speed of 22.2m/s? What would the scale read in this case? (There s an extra page for this one too.) (10pts) θ enter of ircle 100m

1. A sphere with a radius of 1.7 cm has a volume of: A) m 3 B) m 3 C) m 3 D) 0.11 m 3 E) 21 m 3

1. A sphere with a radius of 1.7 cm has a volume of: A) 2.1 10 5 m 3 B) 9.1 10 4 m 3 C) 3.6 10 3 m 3 D) 0.11 m 3 E) 21 m 3 2. A 25-N crate slides down a frictionless incline that is 25 above the horizontal.